1.5 The Nil and Jacobson Radicals - University of Sydney out by the nil and Jacobson radicals yields...
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Transcript of 1.5 The Nil and Jacobson Radicals - University of Sydney out by the nil and Jacobson radicals yields...
1.5 The Nil and Jacobson Radicals
The idea of a “radical” of a ring A is an ideal I
comprising some “nasty” piece of A such that
A/I is well-behaved or tractable.
Two types considered here are the nil and Jacobson
radicals, which are intimately connected with prime
and maximal ideals respectively.
123
Factoring out by the nil and Jacobson
radicals yields quotients closely related to
integral domains and fields respectively (in
a sense which will be made precise).
Consider first all nilpotent elements of a ring:
Theorem: Let A be a ring and put
N = { x ∈ A | x is nilpotent } .
Then N � A and A/N has no nonzero
nilpotent elements.
124
Proof: Let x, y ∈ N , a ∈ A . Then
xm = yn = 0 (∃m,n ∈ Z+) .
Clearly
(ax)m = (−x)m = 0 ,
so
ax , −x ∈ N .
125
Also
(x+ y)m+n−1 =
m+n−1∑
i=0
(
m+ n− 1
i
)
xiym+n−1−i
where(
m+ n− 1
i
)
denotes the usual binomial coefficient, and we
interpret integer multiples in the usual way.
126
If i ≥ m then xi = 0 .
If i < m then
m+ n− 1 − i = n+ (m− i− 1) ≥ n
so that
ym+n−1−i = 0 .
Thus
(x+ y)m+n−1 = 0 ,
so x+y ∈ N , completing the proof that N � A .
127
Further, if N + a is a nilpotent element of A/N
then
(N + a)k = N + ak = N (∃k ∈ Z+)
so ak ∈ N , so
akl = (ak)l = 0 (∃l ∈ Z+) ,
which shows a ∈ N and N + a = N .
Thus all nonzero elements of A/N are not
nilpotent, and the theorem is proved.
128
The ideal N of the previous theorem is called the
nilradical of A , and the last part of the previous
theorem says that
the nilradical of what is left, after factoring
out by the nilradical, is itself trivial.
An alternative description is provided by:
Theorem: The nilradical of a ring is the
intersection of all of the prime ideals of the
ring.
129
Before proving this, we observe that its importance
derives from the following concept:
Suppose { Ai | i ∈ I } is a family of rings and
consider the ring
A =∏
i∈I
Ai = { (xi)i∈I | xi ∈ Ai ∀i }
with coordinatewise operations, called the direct
product of the family.
130
Call a ring B a subdirect product of this family if
there exists a ring homomorphism
ψ : B → A
such that
(i) ψ is injective; and
(ii) ρj ◦ ψ : B → Aj is surjective for all j
where ρj : A → Aj is the projection mapping
(xi)i∈I 7→ xj .
131
Think of subdirect products as being “close” to
direct products.
Observation: Let B be a ring and
{ Ji | i ∈ I } be a family of ideals of B .
Put
J =⋂
i∈I
Ji .
Then J � B and B/J is a subdirect
product of the family { B/Ji | i ∈ I } .
132
Proof of the Observation: The map
ψ : B →∏
i∈I
B/Ji , x 7→ (Ji + x)i∈I
is easily seen to be a ring homomorphism with kernel
kerψ =⋂
i∈I
Ji = J .
Thus
B/J ∼= im ψ .
(by the Fundamental Homomorphism Theorem).
133
Further, clearly the composite
Bψ→
∏
i∈I
B/Jiρj→ B/Jj
is surjective for each j , so
im ψ , and hence B/J ,
is a subdirect product of the family
{ B/Ji | i ∈ I } ,
and the Observation is proved.
134
Corollary to the Theorem: If A is any
ring and N its nilradical, then A/N is a
subdirect product of integral domains.
Proof: Let { Ji | i ∈ I } be the family of all
prime ideals of a ring A . By the Theorem,
N =⋂
i∈I
Ji
is the nilradical of A , and by the Observation, A/N
is a subdirect product of the family { A/Ji | i ∈I } , each member of which is an integral domain.
135
Corollary: A ring with trivial nilradical is
a subdirect product of integral domains.
Proof of the Theorem: Let N denote the
nilradical and N ′ the intersection of all prime ideals
of the ring A .
We have to show N = N ′ .
Consider a prime ideal P of A and let x ∈ N .
136
Then
x(xk−1) = xk = 0 ∈ P (∃k ∈ Z+) ,
so, since P is prime,
x ∈ P or xk−1 ∈ P .
By a simple induction, x ∈ P . Thus N ⊆ P .
Hence
N ⊆ N ′ .
137
Suppose x ∈ A\N , so x is not nilpotent. Put
Σ = { I � A | (∀k ∈ Z+) xk 6∈ I } ,
which is a poset with respect to ⊆ .
Certainly Σ 6= ∅ , since {0} ∈ Σ (because x is
not nilpotent).
It is easy to verify that Zorn’s Lemma
applies, guaranteeing the existence of a
maximal element P of Σ .
138
We will show P is prime.
Suppose a, b ∈ A and ab ∈ P .
Suppose a, b 6∈ P , so
P ⊂ P + aA and P ⊂ P + bA .
But
P + aA , P + bA � A
and P is maximal in Σ . Hence
P + aA , P + bA 6∈ Σ ,
139
so
xk ∈ P + aA , xl ∈ P + bA (∃k, l ∈ Z+) .
Then
xk+l = xk xl ∈ P + abA � A ,
so
P + abA 6∈ Σ .
But ab ∈ P , so P + abA = P ∈ Σ , a
contradiction.
140
Hence a ∈ P or b ∈ P , proving P is prime.
In particular x 6∈ P , so x 6∈ N ′ .
This proves
N ′ ⊆ N ,
so we conclude
N = N ′ ,
and the Theorem is proved.
141
Define the Jacobson radical R of a ring A to be
the intersection of all of the maximal ideals of A .
Thus
A/R is a subdirect product of fields.
Clearly, the maximal ideals of A/R have the form
M/R where M is a maximal ideal of A , so the
Jacobson radical of A/R is just R/R = {R} ,
the trivial ideal.
142
Since all maximal ideals are prime, we get
immediately that
nilradical ⊆ Jacobson radical.
The nilradical was defined in terms of a membership
test for elements (that they be nilpotent). A
membership test exists also for the Jacobson radical.
Theorem: Let x ∈ A . Then
x ∈ R ⇐⇒ (∀y ∈ A) 1 − xy is a unit.
143
Proof: (=⇒) Suppose 1 − xy is not a unit for
some y ∈ A .
By Zorn’s Lemma (an earlier Corollary)
1 − xy ∈ M
for some maximal ideal M .
If x ∈ R then x ∈M
(since R is the intersection of all maximal ideals)
144
and so
1 = (1 − xy) + xy ∈ M ,
elements of M
which is impossible since M 6= A .
Hence x 6∈ R .
145
(⇐=) Suppose x 6∈ R .
Then x 6∈M for some maximal ideal M
(since R is the intersection of all maximal ideals).
By maximality,
A = 〈M ∪ {x}〉
= { m+ xy | m ∈M , y ∈ A } .
146
In particular,
1 = m+ xy (∃m ∈M)(∃y ∈ A)
so
1 − xy = m ∈ M .
If 1 − xy is a unit then M = A .
But M 6= A , so 1 − xy is not a unit.
The Theorem is proved.
147
Examples: (1) In any integral domain the
nilradical is trivial, and in any local ring the Jacobson
radical is the unique maximal ideal.
Since there are local rings which are integral domains
but not fields, the nil and Jacobson radicals need not
be equal.
(2) Let A = Z .
Claim: R = N = {0} .
Proof: Consider 0 6= a ∈ Z .
148
If 1 − 3a = 1 then a = 0 ,
whilst if 1 − 3a = −1 then a = 2/3 ,
both of which are impossible.
Hence 1−3a is not a unit, so a 6∈ R , so R = {0} .
Alternatively (since maximal ideals in Z , being a
PID, are precisely nonzero prime ideals)
R =⋂
p prime
pZ = {0} .
149
(3) Let A = F [x] where F is a field.
Claim: R = N = {0} .
Proof: Consider 0 6= α ∈ F [x] .
Then 1 − xα is a polynomial of degree ≥ 1 , so
cannot be a unit
(units in F [x] being just the nonzero constants).
Hence α 6∈ R , so again R = {0} .
150
(4) Let A = F [G] , a group ring, where F is a
field of prime characteristic p , and G is an abelian
p-group.
Claim: In this example
R = N ={
∑
αg g |∑
αg = 0}
.
Proof: Consider the onto ring homomomorphism
φ : A→ F where∑
αg g 7→∑
αg ,
151
whose kernel is
kerφ = {∑
αg g |∑
αg = 0}
.
Thus
A/ kerφ ∼= F
(by the Fundamental Homomorphism Theorem),
so kerφ is a maximal ideal of A . Hence
R ⊆ kerφ .
152
Consider
x = α1 g1 + . . . + αn gn ∈ kerφ .
Then
α1 + . . . + αn = 0 ,
and, for some sufficiently high power m = pk ,
gmi = 1 (∀i = 1, . . . , n) .
Hence
153
xm = (α1g1 + . . .+ αngn)m
= (α1g1)m + . . .+ (αngn)
m
[by the “Freshman’s Dream”
(a+ b)p = ap + bp
in any field of characteristic p ],
154
so
xm = αm1 gm1 + . . .+ αmn g
mn
= αm1 + . . . + αmn
= (α1 + . . . + αn)m
= 0m = 0 .
155
Thus x ∈ N . Hence R ⊆ kerφ ⊆ N ⊆ R , so
R = N = kerφ ,
and the Claim is proved.
Exercise: Let A be a finite ring.
(1) Prove that if x ∈ A then some power
of x is idempotent.
(2) Verify that if 0 6= e = e2 ∈ A then
1 − e is idempotent so cannot be a unit.
(3) Deduce that N = R .
156